let-f-x-x-x-2-x-1-developp-f-atintegr-serie-

Question Number 35052 by math khazana by abdo last updated on 14/May/18

l e t f (x) = \frac{x}{x^{2} + x - 1}

d e v e l o p p f a t i n t e g r s e r i e

Commented by abdo mathsup 649 cc last updated on 16/May/18

l e t f i n d t h e r o o t s o f x^{2} + x - 1

Δ = 1 - 4 (- 1) = 5 \Rightarrow x_{1} = \frac{- 1 + \sqrt{5}}{2} a n d x_{2} = \frac{- 1 - \sqrt{5}}{2}

f (x) = \frac{x}{(x - x_{1}) (x - x_{2})} = \frac{x}{x_{2} - x_{1}} {\frac{1}{x - x_{2}} - \frac{1}{x - x_{1}}}

= \frac{x}{x_{2} - x_{1}} {\frac{- 1}{x_{2} (1 - \frac{x}{x_{2}})} + \frac{1}{x_{1} (1 - \frac{x}{x_{1}})}} s o f o r

∣ x ∣< i n f (∣ x_{1} ∣, ∣ x_{2} ∣) w e h a v e

f (x) = \frac{x}{- \sqrt{5}} {\frac{1}{x_{1}} \sum_{n = 0}^{\infty} {(\frac{x}{x_{1}})}^{n} - \frac{1}{x_{2}} \sum_{n = 0}^{\infty} {(\frac{x}{x_{2}})}^{n}}

f (x) = \frac{x}{x_{2} \sqrt{5}} \sum_{n = 0}^{\infty} {(\frac{x}{x_{2}})}^{n} - \frac{x}{x_{1} \sqrt{5}} \sum_{n = 0}^{\infty} {(\frac{x}{x_{1}})}^{n} .